Vectors and planes are the longest single chapter on the IB Mathematics Analysis & Approaches HL syllabus and they show up in every Paper 1 and Paper 2 in some form. The topic spans seven sub-syllabus codes (AHL 3.12 through 3.18) and the marks are concentrated in just a handful of recurring traps: forgetting to check the third equation when classifying two lines, using cosine instead of sine for the line-plane angle, citing the perpendicular-distance formula that is not in the IB booklet, and writing "no solution" without naming the geometric configuration of three planes.
This cheatsheet condenses everything from AHL 3.12–3.18 into one page you can revise from. If you want the printable PDF version, the full set of notes, the worked tutorials, and the marked-up solutions, scroll to the bottom for the download links and the gated full library.
§1 — Vectors: Basics & Geometric Proofs AHL 3.12
Core formulas
Magnitude:$|\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$
Unit vector:$\hat{\mathbf{v}} = \dfrac{\mathbf{v}}{|\mathbf{v}|}$, $|\hat{\mathbf{v}}| = 1$
Displacement:$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
Distance:$|AB| = |\mathbf{b} - \mathbf{a}|$
Midpoint:$\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$
Ratio $m:n$:$\overrightarrow{OP} = \dfrac{n\mathbf{a} + m\mathbf{b}}{m + n}$
Parallel:$\mathbf{a} = k\mathbf{b}$, $k \neq 0$
Proof toolkit
| To prove | Show that |
|---|---|
| $PQ \parallel RS$ | $\overrightarrow{PQ} = k\overrightarrow{RS}$ |
| $PQ \perp RS$ | $\overrightarrow{PQ} \cdot \overrightarrow{RS} = 0$ |
| Collinear | $\overrightarrow{AB} = k\overrightarrow{AC}$ AND share a point |
| $|MN| = k|AB|$ | $\overrightarrow{MN} = k\overrightarrow{AB}$ |
| $P$ midpoint of $AB$ | $\overrightarrow{OP} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ |
TrickParallelogram OABC: always $\overrightarrow{AB} = \mathbf{c}$, $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$, $\overrightarrow{CB} = \mathbf{a}$. Set these up before anything else.
TrickRegular hexagon ABCDEF with $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$: $\overrightarrow{OC} = \mathbf{b} - \mathbf{a}$, $\overrightarrow{OD} = -\mathbf{a}$, $\overrightarrow{OE} = -\mathbf{b}$, $\overrightarrow{OF} = \mathbf{a} - \mathbf{b}$.
TrapCollinear requires two things: (1) vectors are parallel, AND (2) they share a point. Missing either loses a mark.
TrapAfter showing $\overrightarrow{MN} = k\overrightarrow{AB}$ you must write: "Since $\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{AB}$, $MN$ is parallel to $AB$." This reasoning sentence earns the R mark.
§2 — Scalar (Dot) Product AHL 3.13
Formulas
Algebraic:$\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$
Geometric:$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$
Angle:$\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}, \;\; \theta \in [0°, 180°]$
Perpendicular:$\mathbf{a} \cdot \mathbf{b} = 0$
Parallel:$|\mathbf{a} \cdot \mathbf{b}| = |\mathbf{a}||\mathbf{b}|$
Self-dot:$\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$
Key identities
Expand:$(\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - |\mathbf{b}|^2$
Perp $\Leftrightarrow$:$|\mathbf{a}|^2 = |\mathbf{b}|^2 \Leftrightarrow (\mathbf{a} + \mathbf{b}) \perp (\mathbf{a} - \mathbf{b})$
Properties:$\mathbf{v} \cdot \mathbf{w} = \mathbf{w} \cdot \mathbf{v}$; $\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w}$
Identity:$|\mathbf{a} \times \mathbf{b}|^2 + (\mathbf{a} \cdot \mathbf{b})^2 = |\mathbf{a}|^2 |\mathbf{b}|^2$
TrickProof shortcut: replace $\mathbf{a} \cdot \mathbf{b}$ with $|\mathbf{a}||\mathbf{b}|\cos\theta$ and $|\mathbf{a} \times \mathbf{b}|$ with $|\mathbf{a}||\mathbf{b}|\sin\theta$, then $\sin^2\theta + \cos^2\theta = 1$ kills the whole thing in two lines.
TrickParallelogram with $|\mathbf{c}| = 2|\mathbf{a}|$: substitute $\mathbf{a} \cdot \mathbf{c} = 2|\mathbf{a}|^2 \cos\theta$ directly — don't try to expand in components.
Note$\theta \in [0°, 180°]$ always — the dot product always gives the unique angle between two vectors, never an ambiguous result.
§3 — Vector (Cross) Product AHL 3.16
Formula & properties
$$\mathbf{a} \times \mathbf{b} = \begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{pmatrix}$$
Magnitude:$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$
Anti-commutative:$\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$
Parallel:$\mathbf{a} \times \mathbf{b} = \mathbf{0}$
Self-cross:$\mathbf{a} \times \mathbf{a} = \mathbf{0}$
Perpendicular:$(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0$ and $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0$
Area applications
Parallelogram:$|\mathbf{a} \times \mathbf{b}|$
Triangle:$\tfrac{1}{2}|\mathbf{a} \times \mathbf{b}|$
Normal to a plane through 3 points: given $A, B, C$ on a plane, $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$. Then plane equation: $\mathbf{n} \cdot \mathbf{r} = \mathbf{n} \cdot \mathbf{a}$.
Simplify first: e.g. $(14, -21, -7) \to 7(2, -3, -1)$. Use $(2, -3, -1)$ as the normal — simpler.
TrickAfter computing $\mathbf{a} \times \mathbf{b}$: immediately verify with $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} \overset{?}{=} 0$ and $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} \overset{?}{=} 0$. Takes 20 seconds and catches all arithmetic errors before they cost you marks.
TrapOrder matters: $\mathbf{a} \times \mathbf{b} \neq \mathbf{b} \times \mathbf{a}$ — the sign flips. Always state which order you used, especially in "show" questions.
§4 — Equations of Lines in 3D AHL 3.14
Three forms — all equivalent
| Form | Equation |
|---|---|
| Vector | $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}, \;\; \lambda \in \mathbb{R}$ |
| Parametric | $x = x_0 + \lambda l$, $y = y_0 + \lambda m$, $z = z_0 + \lambda n$ |
| Cartesian | $\dfrac{x - x_0}{l} = \dfrac{y - y_0}{m} = \dfrac{z - z_0}{n}$ |
Reading a Cartesian line: $\dfrac{x + 1}{2} = y = 3 - z$. Rewrite as $\dfrac{x - (-1)}{2} = \dfrac{y - 0}{1} = \dfrac{z - 3}{-1}$ ⇒ point $(-1, 0, 3)$, direction $(2, 1, -1)$.
Angle between two lines
$$\cos\theta = \frac{|\mathbf{b}_1 \cdot \mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|}$$
Using $|\;\cdot\;|$ in the numerator always gives the acute angle.
Kinematics interpretation
In $\mathbf{r} = \mathbf{a} + t\mathbf{b}$: $t$ is time, $\mathbf{a}$ is the initial position, $\mathbf{b}$ is the velocity vector, $|\mathbf{b}|$ is the speed (include ALL scalar factors!), $\mathbf{r}(t)$ is the position at time $t$.
TrapIf $l = 0$, you cannot write $\dfrac{x - x_0}{0}$ — it is undefined. Instead write $x = x_0$ separately alongside the other two ratios.
TrickKinematics speed: $\mathbf{r} = \mathbf{a} + 4t\mathbf{b}$ has speed $= 4|\mathbf{b}|$, not $|\mathbf{b}|$. The scalar factor before $t$ multiplies the whole velocity vector.
§5 — Classifying Two Lines AHL 3.15
Four possible relationships
| Relationship | Test |
|---|---|
| Identical | Direction parallel + share a point |
| Parallel | $\mathbf{b}_1 = k\mathbf{b}_2$, no shared point |
| Intersecting | Not parallel; 3rd equation consistent |
| Skew | Not parallel; 3rd equation inconsistent |
Skew lines only exist in 3D — not possible in 2D.
The 4-step method
- Check if direction vectors are parallel ($\mathbf{b}_1 = k\mathbf{b}_2$?). If yes: parallel or identical.
- Set $\mathbf{r}_1 = \mathbf{r}_2$ to get 3 equations in $\lambda$ and $\mu$.
- Solve any 2 equations for $\lambda$ and $\mu$.
- Substitute into the 3rd equation. Consistent ⇒ intersecting. Inconsistent ⇒ skew.
TrapThe most common mark lost in this topic: skipping Step 4. Every single time, check the 3rd equation. Solving 2 equations alone is never sufficient to confirm intersection.
TrickTwo ships problem: same position means same value of $t$ for both ships. If you find $\lambda = 2$ from ship A and $\mu = 2$ from ship B, but they used different time variables, it still means they were at that point at different times — check that the time values match.
TrickTo find the point of intersection: once you confirm the 3rd equation is consistent, substitute your $\lambda$ back into $\mathbf{r}_1$ (or $\mu$ into $\mathbf{r}_2$) — they must give the same point.
§6 — Equations of Planes AHL 3.17
Three forms of a plane
| Form | Equation |
|---|---|
| Cartesian | $ax + by + cz = d$, where $\mathbf{n} = (a, b, c)^T$ is the normal |
| Normal / Vector | $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} = d$ |
| Parametric | $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$ ($\mathbf{b}, \mathbf{c}$ lie in the plane, not parallel) |
Finding the equation
From a normal and a point $A$:
- $d = \mathbf{n} \cdot \mathbf{a}$.
- Write $ax + by + cz = d$.
From 3 points $A, B, C$:
- Compute $\overrightarrow{AB}$ and $\overrightarrow{AC}$.
- $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$ (simplify it!).
- $d = \mathbf{n} \cdot \mathbf{a}$.
- Verify all 3 points satisfy the equation.
Parallel & perpendicular
| Relationship | Test |
|---|---|
| $\Pi_1 \parallel \Pi_2$ | $\mathbf{n}_1 = k\mathbf{n}_2$ for some $k$ |
| $\Pi_1 \perp \Pi_2$ | $\mathbf{n}_1 \cdot \mathbf{n}_2 = 0$ |
| Line $\parallel \Pi$ | $\mathbf{b} \cdot \mathbf{n} = 0$ |
| Line $\perp \Pi$ | $\mathbf{b} = k\mathbf{n}$ |
TrapThe perpendicular distance formula is NOT in the IB data booklet. Never try to memorise $d = \dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$. Use the line method: write the line through your point with direction $\mathbf{n}$, find where it meets the plane, then compute the distance. This always works.
TrickVerify a plane equation by checking that all three original points satisfy it. If any point fails, you made an arithmetic error — fix it before moving on.
§7 — Intersections & Angles AHL 3.18
The three angle formulas
| Configuration | Formula |
|---|---|
| Line–Line | $\cos\theta = \dfrac{|\mathbf{b}_1 \cdot \mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|}$ |
| Plane–Plane | $\cos\theta = \dfrac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$ |
| Line–Plane | $\sin\alpha = \dfrac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$ |
$|\;\cdot\;|$ in the numerator always gives the acute angle.
Line meets plane
- Write the line in parametric form: $x = x_0 + \lambda l$, etc.
- Substitute into the plane equation $ax + by + cz = d$.
- Solve for $\lambda$.
- Substitute $\lambda$ back to find the point.
TrapLine–Plane uses $\sin$, not $\cos$. The angle $\alpha$ is measured from the plane surface to the line, not from the normal. The complement of $\alpha$ (i.e. $90° - \alpha$) is the angle between the line and the normal — that one uses $\cos$. This is the #2 most-lost mark in this topic.
Foot of perpendicular from $P$ to plane $\Pi$
- Write the line through $P$ with direction $\mathbf{n}$: $\mathbf{r} = \overrightarrow{OP} + t\mathbf{n}$.
- Substitute into the plane equation, solve for $t$.
- Substitute $t$ back to find the foot $F$.
Reflection of $P$ in plane $\Pi$
Once you have the foot $F$: $P' = 2F - P$. (Because $F$ is the midpoint of $PP'$.)
Two-plane intersection
- Direction of the line of intersection $= \mathbf{n}_1 \times \mathbf{n}_2$.
- Point: set $z = 0$, solve the resulting $2 \times 2$ system.
TrickTwo planes → line: $\mathbf{n}_1 \times \mathbf{n}_2$ gives the direction in one step — far faster than elimination. Then set $z = 0$ (or $x = 0$ if easier) to find one point. Always verify the point lies in both planes.
§8 — Three-Plane Systems AHL 3.18
The 8 cases
| # | Description | Result |
|---|---|---|
| 1 | All three coincident | $\infty$ (plane) |
| 2 | Two coincident, one intersects | $\infty$ (line) |
| 3 | Two coincident, one parallel | No solution |
| 4 | Two parallel, one cuts both | No solution |
| 5 | All three parallel | No solution |
| 6 | All meet at one point | Unique point |
| 7 | Common line (sheaf) | $\infty$ (line) |
| 8 | Triangular prism | No solution |
RREF decision rules
Write the augmented matrix $[A \mid \mathbf{d}]$ and row-reduce:
| Pattern | Conclusion |
|---|---|
| 3 pivot rows, no contradiction | → Case 6 |
| Row $[0 \;\; 0 \;\; 0 \;\mid\; 0]$ | → Case 7 |
| Row $[0 \;\; 0 \;\; 0 \;\mid\; k]$, $k \neq 0$ | → No solution |
| Two zero rows | → Case 1 |
Parametric solution (Case 7): set the free variable $z = t$. Express $x$ and $y$ in terms of $t$. Write as $\mathbf{r} = \mathbf{a} + t\mathbf{d}$.
TrickCheck normals first, before any RREF. If $\mathbf{n}_i = k\mathbf{n}_j$ for any pair, you are in Cases 1–5 immediately — no row reduction needed at all.
Trap"No solution" is never a complete answer. You must name the geometric configuration: triangular prism, two parallel planes, three parallel planes, etc. The IB almost always awards a separate mark for this interpretation.
NoteCases 6–8 occur when all three normals are non-parallel to each other (no two normals are scalar multiples). In this case you must use RREF or elimination to determine which case applies.
§9 — Exam Attack Plan All sections
| Question trigger | Reach for |
|---|---|
| Cartesian line given | Rewrite as $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$, read off point and direction |
| Classify two lines | Step 1: check $\parallel$. Step 2: set equal, solve 2 eqs. Step 3: check 3rd eq. |
| Two lines in 3D, find angle | $\cos\theta = \frac{|\mathbf{b}_1 \cdot \mathbf{b}_2|}{|\mathbf{b}_1||\mathbf{b}_2|}$ |
| Parallel or perp planes | Compare normals: $\mathbf{n}_1 = k\mathbf{n}_2$ (parallel) or $\mathbf{n}_1 \cdot \mathbf{n}_2 = 0$ (perp) |
| Angle between planes | $\cos\theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$ |
| Angle between line and plane | $\sin\alpha = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$ — use $\sin$, not $\cos$ |
| Normal to plane through 3 pts | $\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC}$ |
| Area of triangle / parallelogram | $\tfrac{1}{2}|\mathbf{a} \times \mathbf{b}|$ / $|\mathbf{a} \times \mathbf{b}|$ |
| "Show" involving dot or cross | Use $|\mathbf{a}||\mathbf{b}|\sin\theta$ or $|\mathbf{a}||\mathbf{b}|\cos\theta$ forms, then $\sin^2 + \cos^2 = 1$ |
| Parallelogram OABC proof | Immediately write $\overrightarrow{AB} = \mathbf{c}$, $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$, $\overrightarrow{CB} = \mathbf{a}$ |
| Three-plane system | Check normals first → RREF → name the geometry |
| Ships / planes / helicopters | Speed $= |\text{full velocity}|$. Same position ⇒ same $t$ for both. |
| Line meets plane | Parametric sub into plane eq, solve $\lambda$, find point |
| Foot of perp from $P$ to $\Pi$ | Line through $P$ with dir $\mathbf{n}$, sub into $\Pi$, solve $t$, find $F$ |
| Reflection of $P$ in plane | $P' = 2F - P$ where $F$ is the foot of perpendicular |
| Two planes → line | Direction $= \mathbf{n}_1 \times \mathbf{n}_2$, then set $z = 0$ for a point |
Top 5 marks lost — based on 37 past-paper questions
- Skipping the 3rd equation check when classifying lines. You must check all three equations. Solving just two is not enough to confirm intersection.
- Using $\cos$ instead of $\sin$ for line–plane angle. Line–line and plane–plane use $\cos$. Line–plane uses $\sin$. The reason: $\alpha$ is measured from the surface, not from the normal.
- Missing the geometric interpretation in three-plane systems. After RREF, name the result: "triangular prism", "sheaf of planes meeting in a line", etc. This mark is awarded separately from the algebra.
- Using the perpendicular distance formula that is not in the booklet. $d = \dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$ does NOT appear in the IB formula booklet. Use the line method every time.
- No conclusion sentence in geometric proofs. After showing $\overrightarrow{MN} = k\overrightarrow{AB}$, you must write: "Since $\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{AB}$, $[MN]$ is parallel to $[AB]$." The calculation alone does not earn the reasoning mark.
Worked Example — IB-Style Classifying Two Lines
Question (HL Paper 1 style — 7 marks)
Consider the lines $L_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix} 4 \\ 0 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. Determine whether $L_1$ and $L_2$ intersect, are parallel, or are skew. If they intersect, find the point of intersection.
Solution
- Test for parallel: $\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \neq k\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ for any scalar $k$ (the ratios $2/1, -1/1, 1/-1$ differ). So $L_1$ and $L_2$ are not parallel. (M1)
- Set $\mathbf{r}_1 = \mathbf{r}_2$:
$1 + 2\lambda = 4 + \mu \;\;\;\;\; (i)$
$2 - \lambda = \mu \;\;\;\;\;\;\;\;\;\;\;\;\;\; (ii)$
$3 + \lambda = 5 - \mu \;\;\;\;\;\; (iii)$ (M1) - Solve (i) and (ii): sub (ii) into (i): $1 + 2\lambda = 4 + (2 - \lambda) = 6 - \lambda \Rightarrow 3\lambda = 5 \Rightarrow \lambda = \tfrac{5}{3}$. Then $\mu = 2 - \tfrac{5}{3} = \tfrac{1}{3}$. (A1)(A1)
- Substitute into the 3rd equation: LHS $= 3 + \tfrac{5}{3} = \tfrac{14}{3}$; RHS $= 5 - \tfrac{1}{3} = \tfrac{14}{3}$. The 3rd equation is consistent ⇒ the lines intersect. (R1)
- Find the point of intersection: sub $\lambda = \tfrac{5}{3}$ into $L_1$: $\mathbf{r} = \begin{pmatrix} 1 + \tfrac{10}{3} \\ 2 - \tfrac{5}{3} \\ 3 + \tfrac{5}{3} \end{pmatrix} = \begin{pmatrix} \tfrac{13}{3} \\ \tfrac{1}{3} \\ \tfrac{14}{3} \end{pmatrix}$. (Verify with $\mu = \tfrac{1}{3}$ in $L_2$: $\begin{pmatrix} 4 + \tfrac{1}{3} \\ \tfrac{1}{3} \\ 5 - \tfrac{1}{3} \end{pmatrix} = \begin{pmatrix} \tfrac{13}{3} \\ \tfrac{1}{3} \\ \tfrac{14}{3} \end{pmatrix}$.) (A1)(A1)
Examiner's note: the R1 in step 4 is for the explicit consistency check on the third equation. Solving only two equations and stating "the lines intersect at …" without verifying the third equation loses the R1 every time. Always write a sentence stating that the third equation is satisfied (or not) before drawing your conclusion.
Common Student Questions
Why is the angle between a line and a plane $\sin$, not $\cos$?
For line–line and plane–plane angles you compare two direction or normal vectors directly, so $\cos\theta = \dfrac{|\mathbf{a} \cdot \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}$ gives the acute angle between them. For a line and a plane the angle $\alpha$ is measured from the plane surface to the line — that is the complement of the angle between the line and the normal. Using $\cos$ gives the wrong angle ($90° - \alpha$). The correct formula is $\sin\alpha = \dfrac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}||\mathbf{n}|}$. This is the second most lost mark in the topic.
Do I really need to check the third equation when classifying two lines?
Yes — it is the single most common mark loss in the entire vectors topic. Solving any two of the three equations always gives some $\lambda$ and $\mu$, but only the third equation tells you whether those values are consistent. Consistent means the lines intersect; inconsistent means they are skew. Skipping this step makes every skew lines question look like an intersection.
What goes wrong with the perpendicular distance formula?
The formula $d = \dfrac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$ is not in the IB AA HL formula booklet. If you cite it without derivation, you can lose marks. The reliable IB-approved approach is the line method: write the line through the point with direction vector $\mathbf{n}$, find where it meets the plane (solve for the parameter), and compute the distance from the point to that intersection.
How do I know whether three planes form a triangular prism or a sheaf?
Row-reduce the augmented matrix $[A \mid \mathbf{d}]$. If you get a row of the form $[0 \;\; 0 \;\; 0 \;\mid\; 0]$ (a redundant equation), the planes share a common line (sheaf) — infinitely many solutions. If you get $[0 \;\; 0 \;\; 0 \;\mid\; k]$ with $k \neq 0$ (a contradiction), there is no common point and the configuration is a triangular prism. Stating only "no solution" loses the geometric-interpretation mark — name the configuration explicitly.
How do I find the speed in a vector kinematics problem?
In $\mathbf{r}(t) = \mathbf{a} + t\mathbf{b}$, the velocity vector is $\mathbf{b}$ and the speed is $|\mathbf{b}|$. But if the equation is $\mathbf{r}(t) = \mathbf{a} + 4t\mathbf{b}$, the velocity is $4\mathbf{b}$ and the speed is $4|\mathbf{b}|$ — every scalar in front of $t$ multiplies the velocity. The most common error is reading off $|\mathbf{b}|$ while ignoring the scalar. Always factor out scalars from the $t$-term before computing speed.
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